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We are given two integers a and b, such that, 2 < a < b and b is not a
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01 Apr 2013, 01:05
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39% (02:33) correct 61% (02:30) wrong based on 164 sessions
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We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1? (1) The least common multiple of a and b is 42 (2) The greatest common factor of a and b is 2
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Re: We are given two integers a and b, such that, 2 < a < b and b is not a
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01 Apr 2013, 01:54
guerrero25 wrote: We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1? (1) The least common multiple of a and b is 42 (2) The greatest common factor of a and b is 2 I am struggling in this type of questions ..Any help is greatly appreciated. thanks! From F.S 1 , for a=6,b=7, we have the remainder as 1, which is not more than 1. Again, for a=3,b=14, we have a remainder as 2, which is more than 1. Insufficient. From F.S 2, the integers a and b are of the form 2x and 2y, where x and y are coprimes. Also, 2y = 2x*q + R, where q is a nonnegative integer. or R = 2(yqx). As x and y are coprimes, thus y is not equal to qx, for any integral value of q. Thus, (yqx) will never be zero. And as R is always positive, the value of R will always be more than 1. Sufficient. Note that as a>2, we can have the first value of a only as 4. B.
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Re: We are given two integers a and b, such that, 2 < a < b and b is not a
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01 Apr 2013, 02:03
guerrero25 wrote: We are given two integers a and b, such that, 2 < a < b and b is not a multiple of a. Is the remainder of the division of b by a greater than 1? (1) The least common multiple of a and b is 42 (2) The greatest common factor of a and b is 2 I am struggling in this type of questions ..Any help is greatly appreciated. thanks! STAT1 is not sufficient as we can have mutiple cases case 1 a=6, b=7 now remainder when b is divided by a will be 1 which is NOT greater than 1 case 2 a=6, b=14 now reaminder when b is divided by a will be 2 which is greater than 1 So, STAT1 is NOT sufficient STAT2 GCD of a and b is 2 means that both a and b are even numbers And we know that b is not a mutiple of a so, in any case reamindder when b is divided by a will be more than 1 case1 a=4, b=6 remaidner will be 2 greater than 1 case 2 a=6, b=10 reaminder will be 4 greater than 1 Also, if you notice then the reaminder is a even number greater than or equal to 2 So STAT2 is SUFFICIENT hence, answer will be B Hope it helps!
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Re: We are given two integers a and b, such that, 2 < a < b and b is not a
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01 Apr 2014, 06:26
Alright, as Hamilton Lin says, let's do it!
Statement 1 tells us that the LCM of (a,b) is 42. Therefore, we have that b=7,a=6 remainder is 1, but if b=14 and a=3 remainder is 2, therefore two different answers–> Insufficient.
Statement 2, says that the GCF of (a,b) is 2. Therefore since both are even and b is not a multiple of a the remainder will always be 2, therefore Sufficient
B is the correct answer
Cheers J



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Re: We are given two integers a and b, such that, 2 < a < b and b is not a
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21 Mar 2018, 01:19
Solution: We are given: • ‘\(b\)’ is not a multiple of ‘\(a\)’.
• Both, ‘\(a\)’ and ‘\(b\)’, are greater than \(2\) and ‘\(b\)’ is greater than ‘\(a\)’. Statement1 is “The least common multiple of ‘a’ and ‘b’ is 42”.\(42= 2*3*7\) ‘\(a\)’ and ‘\(b\)’ can have multiple values based on prime factorization of \(42\). Since the value of remainder may or may not greater than 1. Hence, Statement 1 alone is not sufficient to answer the question. Statement2 is “The greatest common factor of ‘a’ and ‘b’ is 2.”Thus, • \(a=2x\) and \(b=2y\), where \(x\) and \(y\) are coprime numbers. Since ‘\(b\)’ is greater than ‘\(a\)’, ‘\(b\)’ on dividing by ‘\(a\)’ can be written as: • \(2y= 2x*m + r\), where ‘\(r\)’ is the remainder and ‘\(r\)’ cannot be ‘\(0\)’ as ‘\(b\)’ is not a multiple of ‘\(a\)’.
• Since \(2y\) is an even number, \((2x*m + r)\) should be an even number.
o ‘\(r\)’ must be even.
Thus, the least value of ‘\(r\)’ can be \(2\). Hence, Statement 2 alone is sufficient to answer the question. Answer: B
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Re: We are given two integers a and b, such that, 2 < a < b and b is not a
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21 Mar 2018, 01:19






